Olympiad Algebra II: Polynomials, Recurrences & Radical Equations

Sums $6, 11, 6$ give $(x-1)(x-2)(x-3)$, whose value at $4$ is $6$; $x + y = 4$, $x^3 + y^3 = 28$ forces $xy = 3$ and $x^2 + y^2 = 10$. Every "find this symmetric expression" problem is a two-step: reach $e_1, e_2, (e_3)$, then rebuild. If the problem wants the actual roots, factor the rebuilt polynomial by testing small integers (rational root theorem).

$a_{n+2} = a_{n+1} + 2a_n$ has roots $2, -1$: $a_n = \tfrac{2^n - (-1)^n}{3}$, so $a_{10} = 341$. The method IS diagonalization of the companion matrix (Linear Algebra course), which is why it also handles $k$-term recurrences and why a root of modulus $> 1$ dominates growth. Nonhomogeneous recurrences add a particular solution, exactly as for differential equations.

$\sqrt{x + 5} = x - 1$ squares to $(x - 4)(x + 1) = 0$; $x = -1$ is a ghost (the left side is $2$, the right $-2$). The clean discipline: record the domain and sign constraints BEFORE squaring ($x - 1 \ge 0$ here), then squaring is reversible on that domain. For nested radicals, isolate one radical per squaring; for $\sqrt{a} \pm \sqrt{b}$ forms, multiply by the conjugate.

Divide by $x^2$: $\left(x^2 + \tfrac{1}{x^2}\right) - 5\left(x + \tfrac1x\right) + 6 = 0$. With $t = x + \tfrac1x$, $x^2 + \tfrac1{x^2} = t^2 - 2$: $t^2 - 5t + 4 = 0$, $t = 1$ or $4$. $t = 4$ gives $x = 2 \pm \sqrt3$; $t = 1$ gives complex roots. Real solutions: $2 \pm \sqrt3$.

Both $\alpha^n$ and $\beta^n$ satisfy the recurrence (substitute and factor out $r^n$). Solutions form a vector space closed under linear combinations, and a solution is determined by $(a_0, a_1)$ — two numbers — so the space is 2-dimensional. $\alpha^n$ and $\beta^n$ are independent (their initial pairs $(1, \alpha)$, $(1, \beta)$ are independent for $\alpha \ne \beta$), hence a basis. $\blacksquare$ For a double root, $n\alpha^n$ is the second basis element — verify by substitution, exactly as for $xe^{rx}$ in differential equations.

Multiply $\sum a_k(p/q)^k = 0$ by $q^n$: $\sum a_kp^kq^{n-k} = 0$. Every term except $a_0q^n$ is divisible by $p$, so $p \mid a_0q^n$, and $\gcd(p, q) = 1$ forces $p \mid a_0$. Symmetrically every term except $a_np^n$ is divisible by $q$, so $q \mid a_n$. $\blacksquare$ Significance: it makes "factor by finding an integer root" a finite search, and its special case "a monic integer polynomial has no non-integer rational roots" is why $\sqrt2$ is irrational.

Step 1 — $e_1 = 6$; $e_2 = \tfrac{e_1^2 - p_2}{2} = \tfrac{36 - 14}{2} = 11$; Newton: $p_3 = e_1p_2 - e_2p_1 + 3e_3$ gives $36 = 84 - 66 + 3e_3$, so $e_3 = 6$.

Step 2 — the triple consists of the roots of $t^3 - 6t^2 + 11t - 6 = (t-1)(t-2)(t-3)$.

Step 3 — all permutations of $(1, 2, 3)$: six solutions. Symmetric systems never require solving for one variable at a time; rebuild the polynomial and factor.

Step 1 — the fixed point: $L = 2L + 3$ gives $L = -3$.

Step 2 — shift: $b_n = a_n + 3$ satisfies $b_{n+1} = 2b_n$, so $b_n = 2^nb_0 = 4\cdot2^n$.

Step 3 — $a_n = 4\cdot2^n - 3 = 2^{n+2} - 3$; $a_{10} = 4096 - 3 = 4093$.

Step 4 — check: $a_1 = 2 + 3 = 5 = 8 - 3$ ✓. Subtracting the fixed point is the recurrence version of "particular solution plus homogeneous," and it works for any $a_{n+1} = ka_n + c$ with $k \ne 1$.